Question Video: Finding the Absolute Maximum and Minimum Values of a Function in a Given Interval Involving Using the Product Rule | Nagwa Question Video: Finding the Absolute Maximum and Minimum Values of a Function in a Given Interval Involving Using the Product Rule | Nagwa

# Question Video: Finding the Absolute Maximum and Minimum Values of a Function in a Given Interval Involving Using the Product Rule Mathematics • Third Year of Secondary School

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Find the absolute maximum and minimum values rounded to two decimal places of the function π(π₯) = 5π₯π^(βπ₯), π₯ β [0, 4]

02:08

### Video Transcript

Find the absolute maximum and minimum values rounded to two decimal places of the function π of π₯ equals five π₯π to the negative π₯, given that π₯ is a part of the closed interval zero to four.

Remember, to find absolute extrema for our function π of π₯, we follow three steps. We find all critical points in our closed interval. We then find the values of π of π₯ at these critical points. And then, we check the end points for absolute extrema. The critical points are the points where the derivative is equal to zero or does not exist. So, weβre going to find the derivative of our function and set that equal to zero.

And here, we notice that this itself is the product of two differentiable functions. So, weβre going to use the product rule. This says that the derivative of the product of two differentiable functions π’ and π£ is π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯. So, we let π’ be equal to five π₯ and π£ be equal to π to the negative π₯. Then, dπ’ by dπ₯ is equal to five and dπ£ by dπ₯ π₯ is equal to negative π to the negative π₯. This means the derivative of our function is five π₯ times negative π to the negative π₯ plus π to the negative π₯ times five, which we can simplify to five π to the negative π₯ times one minus six

Letβs set this equal to zero. Now thereβs no way for five π to the negative π₯ to be equal to zero. So, for the statement five π to the negative π₯ times one minus π₯ equals zero to be true, we know that one minus π₯ itself must be equal to zero, which means that π₯ equals one is a critical point. Weβre, therefore, going to evaluate our function at this critical point and at the end points of the function, so π of one, π of zero, and π of four.

π of one is five times one times π to the negative one, which is 1.8393 and so on, or correct to two decimal places as required 1.84. π of zero is zero. And π of four is five times four times π to the negative four, which is 0.37 correct to two decimal places. And we can, therefore, say that the absolute maximum value of our function is 1.84 and the absolute minimum value is zero.

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